/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 Differentiate each function. $... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 43

Differentiate each function. $$F(t)=\frac{1}{t-4}$$

Short Answer

Expert verified
- \( \frac{1}{(t-4)^{2}} \)

Step by step solution

01

- Rewrite the function using exponent notation

Rewrite the function to make differentiation easier. Start by expressing the denominator as a negative exponent: \[F(t) = \frac{1}{t-4} = (t-4)^{-1}\]
02

- Apply the power rule

Use the power rule for differentiation, which states that \(\frac{d}{dx} x^n = n x^{n-1}\). Applying this to \( (t-4)^{-1} \), we get: \[ \frac{d}{dt} (t-4)^{-1} = -1 (t-4)^{-2} \]
03

- Use the chain rule

Finally, apply the chain rule, which states that \(\frac{d}{dx} f(g(x)) = f'(g(x)) \times g'(x)\). Here, we have \( f(g(t)) = (g(t))^{-1} \) with \( g(t) = t-4 \): \[ \frac{d}{dt} (t-4)^{-1} = -1 (t-4)^{-2} \times \frac{d}{dt} (t-4) = -1 (t-4)^{-2} \times 1 = - (t-4)^{-2} \]
04

- Simplify the result

Rewrite the result from exponent notation back to fractional notation: \[ - (t-4)^{-2} = - \frac{1}{(t-4)^{2}} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

power rule
The power rule is a fundamental technique in differentiation. It allows us to find the derivative of a function of the form \( x^n \). The power rule states: \[ \frac{d}{dx} x^n = n x^{n-1} \].
This rule is essential for simplifying the differentiation process, especially when dealing with polynomial expressions or functions raised to a power.
chain rule
The chain rule is another crucial tool in differentiation. It is used when differentiating a composite function, which is a function nested within another function. The chain rule states: \[ \frac{d}{dx} f(g(x)) = f'(g(x)) \times g'(x) \].
The chain rule is particularly useful when you encounter functions where one variable depends on another. In this problem, we applied the chain rule to \( (t-4)^{-1} \) by identifying \( g(t) = t-4 \). After differentiating the outer function, we multiplied by the derivative of the inner function, which was \(\frac{d}{dt}(t-4) = 1 \).
negative exponent
Understanding negative exponents can make differentiation easier. A negative exponent \( x^{-n} \) is defined as the reciprocal of the positive exponent: \[ x^{-n} = \frac{1}{x^n} \].
This principle allows us to rewrite functions with negative exponents in a more straightforward form for differentiation. For example, in our problem, we rewrote the fraction \( \frac{1}{t-4} \) as \( (t-4)^{-1} \). This simplification prepared us to apply the power rule and chain rule effectively.

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Most popular questions from this chapter

Consumer demand. Suppose that the demand function for a product is given by \(D(p)=\frac{80,000}{p}\) and that price \(p\) is a function of time given by \(p=1.6 t+9,\) where \(t\) is in days. a) Find the demand as a function of time \(t\) b) Find the rate of change of the quantity demanded when \(t=100\) days.

Find the points on the graph of \(y=2 x^{6}-x^{4}-2\) at which the tangent line is horizontal.

The average price, in dollars, of a ticket for a Major League baseball game \(x\) years after 1990 can be estimated by \(p(x)=9.41-0.19 x+0.09 x^{2}\) a) Find the rate of change of the average ticket price with respect to the year, \(d p / d x\). b) What is the average ticket price in \(2010 ?\) c) What is the rate of change of the average ticket price in \(2010 ?\)

Find \(d y / d x .\) Each function can be differentiated using the rules developed in this section, but some algebra may be required beforehand. $$y=\sqrt[3]{8 x}$$

Find \(d y / d x .\) Each function can be differentiated using the rules developed in this section, but some algebra may be required beforehand. $$y=\frac{x^{5}-3 x^{4}+2 x+4}{x^{2}}$$
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